I have the following matrix:

l = {{{"black_pepper", "cabbage", "carrot", "cumin", "olive_oil", 
 "onion", "potato", "TURMERIC"}, {"bell_pepper", "cayenne", 
 "cilantro", "cumin", "garlic", "lemon_juice", "olive_oil", 
 "parsley"}, {"butter", "cayenne", "coconut", "onion", 
 "roasted_peanut", "tomato", "vegetable_oil", "vinegar"}}};

I am trying to select the element that has "TURMERIC" in it. So the output should be the first element of l. I tried:

Select[l, StringContainsQ@# == "TURMERIC" &]

but it returns empty list. What am I doing wrong here?

  • $\begingroup$ You have a list of lists of strings, so you need a selection function that acts on a list. Try Select[l, MemberQ[#, StringContainsQ[#, "TURMERIC"] &] &]. (Can't test since I only have gedanken Mathematica currently.) $\endgroup$ Oct 9, 2018 at 15:52
  • $\begingroup$ @J.M.issomewhatokay. Thanks but when I run your code, it returns empty list as well. $\endgroup$
    – Wiliam
    Oct 9, 2018 at 15:54
  • $\begingroup$ Hmm, sorry; try this instead: Select[l, MemberQ[#, s_String /; StringContainsQ[s, "TURMERIC"]] &] $\endgroup$ Oct 9, 2018 at 15:55
  • $\begingroup$ Select[Catenate@l, MemberQ[#, "TURMERIC"] &] or Select[Catenate@l, MemberQ[#, "TURMERIC"] &]//Flatten (or Cases[l, {___, "TURMERIC", ___}, -1]) $\endgroup$
    – user1066
    Oct 9, 2018 at 19:39

1 Answer 1


There is a somewhat superfluous first level in l, so we map the operator form of select over it. Moreover, we just need MemberQ, not StringContainsQ, no?

Select[MemberQ["TURMERIC"]] /@ l

{{{"black_pepper", "cabbage", "carrot", "cumin", "olive_oil", "onion", "potato", "TURMERIC"}}}

  • $\begingroup$ Thanks Select[l, MemberQ[#, s_String /; StringContainsQ[s, "TURMERIC"]] &] works if there is no {{{. I need the out put to be the whole element though. If you read the question, I have said I needed the elements to be in the format: {"black_pepper", "cabbage", "carrot", "cumin", "olive_oil", "onion", "potato", "TURMERIC"} $\endgroup$
    – Wiliam
    Oct 9, 2018 at 16:04

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